Question

Let $f$ be a differentiable function satisfying $f(x)=\frac{2}{\sqrt{3}} \int\limits_0^{\sqrt{3}} f\left(\frac{\lambda^2 x}{3}\right) d \lambda, x>0$ and $f(1)=\sqrt{3}$. If $y=f(x)$ passes through the point $(\alpha, 6)$, then $\alpha$ is equal to_______

Answer 1

Let, $\frac{\lambda^2 x }{3}= t$
$ \Rightarrow \frac{2 \lambda x }{3} d \lambda= dt$
$ \Rightarrow d \lambda=\frac{3}{2} \cdot \frac{1 \sqrt{ x }}{ x \cdot \sqrt{3} \sqrt{ t }} dt$
$ \Rightarrow d \lambda=\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{ x }} \cdot \frac{ dt }{\sqrt{ t }}$
So, $f(x)=\frac{1}{\sqrt{x}} \int_0^x \frac{f(t)}{\sqrt{t}} d t$
$\Rightarrow \sqrt{ x } \cdot f ^{\prime}( x )+\frac{ f ( x )}{2 \sqrt{ x }}=\frac{ f ( x )}{\sqrt{ x }}$
$ \Rightarrow \sqrt{ x } \cdot f ^{\prime}( x )=\frac{ f ( x )}{2 \sqrt{ x }} $
$ \Rightarrow \frac{ dy }{ y }=\frac{ dx }{2 x }$
$\Rightarrow \ln y=\frac{1}{2} \ln x+c \Rightarrow f(x)=\sqrt{x} $
$ \Rightarrow y=\sqrt{3 x} \{\text { as } f(1)=\sqrt{3}\}$
So, $f(x)=\sqrt{3 x}$
Now, $f (\alpha)=6 \Rightarrow 36=3 \alpha$
$\Rightarrow \alpha=12$